- Cute! (but are they really!?)
- Furry! (guess how may hairs per in2? … (hint, humans are born with ~100,000 total)
- 150,000 hairs / cm2 (>1,000,000 / in2)
11/22/2021
(Estes et al. 1974)
300,000 in 1740 … < 2,000 in 1900.
… the rush for the otters’ “soft gold” was a predictable boom and bust cycle, a cautionary example of unsustainable resource use, and a socioeconomic driver of Western—mainly American—involvement in the Pacific region starting in the eighteenth century. (Loshbaugh 2021)
art by John Livingston
Remnant populations from Aleutian Islands … released in OR, WA, BC and SE-AK 1969 – 1972.
\[\Large N_t = N_{t - \Delta t} + \Delta N\]
slight rearrangement:
\[\Large N_{t+1} = N_t + \Delta N\] For now, we just \(\Delta t = 1\), i.e. it’s the discrete unit that we measure population change. VERY TYPICALLY - whether because of biology or field seasons: \[\Delta t = 1\,\, \textrm{year}\].
This is a closed population … and what we will be (mainly) dealing with for the next 3 weeks.
The number of Births and Deaths is proportional to N.
\[\Large N_{t+1} = N_t + bN_t - dN_t\] What does that mean?
Redefine \(\lambda = b - d:\)
\[N_{t+1} = N_t + r_0 N_t\] \[N_{t+1} = (1 + r_0) N_t\]
\(r_0\) intrinsic growth, i.e. proportion increase per unit time).
\[N_{t+1} = \lambda N_t\]
\[N_{t+1} = \lambda(N_t)\] \[N_{t+2} = \lambda(N_{t+1}) = \lambda^2 N_t\] \[N_{t+3} = \lambda^3 N_t\]
Solution:
\[\large N_{t+y} = \lambda^y N_t\]Let’s do some trickery, starting with: \[N_{t+1} = (1 + r_0) N_t\] \[N_{t+1} - N_t = r_0 N_t\] \[N_{t+\Delta t} - N_t = r_{\Delta t} N_t\]
\[\lim_{\Delta t \to 0} {\Delta N \over \Delta t} = \lim_{\Delta t \to 0} {r_{\Delta t} \over \Delta t} N\]
Magically define: \({r_\Delta \over \Delta t} = r\) and rewrite \(\Delta\) as \(d\):
\[\large {dN \over dt} = r N\]
Calculate:
\(\begin{align} {1\over N} dN &= r dt \\ \int_{t' = t_0}^t {1 \over N(t)} dN &= \int_{t' = t_0}^t r dt \\ \log(N) &= rt + C_0 \\ N &= e^{rt + C_0} \\ \end{align}\)
Solution:
Plug in: \[N(0) = N_0\\\] \[ N(t) = N_0 e^{rt}\]
Difference equations \[N_{t+\Delta t} - N_t = \lambda_{\Delta t} N_t\]
think of absolute change
Pros:
Cons:
Differential equations \[{dN \over dt} = r N\]
think of rates (change/time).
Pros
Cons
Source:
https://wdfw.wa.gov/species-habitats/species/enhydra-lutris-kenyoni#resources
Load data:
## year count ## 1 1970 59 ## 2 1989 208 ## 3 1990 212 ## 4 1991 276 ## 5 1992 313 ## 6 1993 307
## ## Call: ## lm(formula = log(count) ~ I(year - 1970), data = WA) ## ## Residuals: ## Min 1Q Median 3Q Max ## -0.191084 -0.062944 -0.005104 0.055518 0.231704 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 4.082641 0.073024 55.91 <2e-16 *** ## I(year - 1970) 0.073251 0.002367 30.95 <2e-16 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.1094 on 23 degrees of freedom ## Multiple R-squared: 0.9766, Adjusted R-squared: 0.9755 ## F-statistic: 958.1 on 1 and 23 DF, p-value: < 2.2e-16
\[\log(N_i) = \alpha + \beta \, Y_i\] \[N_i = \exp(\alpha) \times \exp(\beta \, Y_i)\] \[N_i = e^\alpha {e^\beta}^{Y_i}\] \[N_i = N_0 \lambda ^ {Y_i}\]
where \(N_0 = e^{\alpha} = e^{4.08} = 59.14\), and \(\lambda = e^{\beta} = e^{0.07325} = 1.076\).
SO … percent rate of growth is about 7.6%!
There is no exception to the rule that every organic being increases at so high a rate, that if not destroyed, the earth would soon be covered by the progeny of a single pair … The elephant is reckoned to be the slowest breeder of all known animals, and I have taken some pains to estimate its probable minimum rate of natural increase: it will be under the mark to assume that it breeds when thirty years old, and goes on breeding till ninety years old, bringing forth three pairs of young in this interval; if this be so, at the end of the fifth century there would be alive fifteen million elephants, descended from the first pair.
Charles Darwin - Origin of Species